Solution

1. EXAMPLE 3.1 Number of Neutrons 23592 How many neutrons are there in the U nucleus? Solution Simply subtract the atomic number Z (number of protons) from the nucleon number A (number of protons plus neutrons). A–Z= number of neutrons 235 – 92 = 143 There are 143 neutrons in the nucleus. Exercise 3.1A Exercise 3.1B How many neutrons are there in the I nucleus? A potassium isotope has 21 neutrons in its nucleus. What is the nucleon number and name of the isotope? 131 53

2. EXAMPLE 3.2 Number of Neutrons Solution From the periodic table on the inside front cover, we find the atomic number of strontium is 38. The nucleon number is given as 90. The number of neutrons is therefore 90 – 38 = 52 Exercise 3.2A Exercise 3.2B How many neutrons are there in the molybdenum-90 nucleus? Use the X notation to represent the isotope of uranium having 148 neutrons. A Z How many neutrons are there in the strontium-90 nucleus?

3. CONCEPTUAL EXAMPLE 3.3 Isotopes Refer to the following isotope symbols, in which we use the letter X as the symbol for all elements so that the symbol will not identify the elements. (a) Which are isotopes of the same element? (b) Which have the same nucleon number? (c) Which have the same number of neutrons? X X X X X 16 8 16 7 14 7 14 6 12 6 Solution X X 14 7 16 7 X X 14 6 12 6 Exercise 3.3B X X 16 8 16 7 How many different elements are represented in Exercise 3.3A? X X 14 7 14 6 X X 16 8 14 6 Exercise 3.3A Which of the following are isotopes of the same element? X X X X X 90 37 90 35 88 37 88 38 93 38 Isotopes of the same element will have the same atomic number (subscript). Therefore, and are isotopes of nitrogen (N), and and are isotopes of carbon (C). The nucleon number is the superscript, so and have the same nucleon number. The first is an isotope of oxygen, and the second an isotope of nitrogen. and also have the same nucleon number. The first is an isotope of nitrogen, and the second an isotope of carbon. To determine the number of neutrons, we subtract the atomic number from the nucleon number. We find that and each have eight neutrons (16 – 8 = 8 and 14 – 6 = 8 respectively).

4. EXAMPLE 3.4 Electron Shell Capacity Solution The maximum number of electrons in a given shell is given by 2n2. For the fifth level, n = 5, and so we have 2 x 52 = 2 x 25 = 50 Exercise 3.4A Exercise 3.4B What is the maximum number of electrons in the fourth shell (fourth energy level). What is the lowest shell (value of n) that can hold 18 electrons? What is the maximum number of electrons in the fifth shell (fifth energy level)?

5. EXAMPLE 3.5 Main-Shell Electron Configurations Solution a. Fluorine has the symbol F (Z = 9) it has nine electrons. Two of these electrons go into the first shell, and the remaining seven go into the second level. F 2 7 b. Aluminum (Z = 13) has 13 electrons. Two go into the first shell, eight go into the second, and the remaining three go into the third shell. Al 2 8 3 Notice that fluorine is in group 7A of the periodic table and that it has seven outer-shell electrons. Aluminum is in group 3A, and it has three electrons in its outermost shell. Exercise 3.5 Give the main-shell electron configurations for (a) beryllium and (b) magnesium. These elements are in the same group of the periodic table. What do you notice about the number of electrons in their outermost shells? What are the main-shell electron configurations for (a) fluorine and (b) aluminum?

6. EXAMPLE 3.6 Subshell Notation Solution a. Oxygen (Z = 8) has eight elections. Place them in the lowest unfilled energy sublevels. Two go into the 1s orbital and two into the 2s orbital. That leaves four electrons to be placed in the 2p subshell. The electron configuration is 1s22s22p4. Sulfur atoms have 16 electrons each. The electron configuration is 1s22s22p63s23p4. Note that the total of the superscripts is 16 and that we have not exceeded the maximum capacity for any sublevel. Both O and S have electron configurations with four electrons in their highest energy sublevel (outermost subshell). Exercise 3.6A Exercise 3.6B Without referring to Table 3.3, use subshell notation to write out the electron configurations for (a) fluorine and (b) chlorine. What similarity of features do the electron configurations exhibit? Use Figure 3.16 to write the electron configurations for (a) titanium (Ti) and (b) gallium (Ga). Without referring to Table 3.3, use subshell notation to write out the electron configuration for (a) oxygen and (b) sulfur. What similarity of features do the electron configurations exhibit?

7. EXAMPLE 3.7 Valence Shell Electron Configurations Solution a. Strontium is in group 2A and thus has two valence electrons in an s subshell. Because strontium is in the fifth period of the periodic chart, its outer main shell is n = 5. Its outer electron configuration is therefore 5s2. b. Arsenic is in group 5A and the fourth period. Its five outer (valence) electrons are in the n = 4 shell and have the configuration 4s24p3. Exercise 3.7A Exercise 3.7B Write out the subshell notation for the electrons in the highest main shell for (a) rubidium (Rb), (b) selenium (Se), and (c) germanium (Ge). The valence electrons in aluminum have the configuration 3s23p1. Gallium is directly below aluminum, and indium is directly below gallium in the periodic table. Use only this information to write the configuration for the valence electrons in (a) gallium and (b) indium. Write out the subshell notation for the electrons in the highest main shell for (a) strontium (Sr) and (b) arsenic (As).